Book has the following & solution to it too, pls clear my confusion:
On rainy day , five gentlemen A, B, C,D, E attend a party after leaving their umbrellas in a checkroom. After the party is over, the umbrellas get mixed up and are returned to the gentlemen in such a manner that none receives his own umbrellas. In how many ways can this happen? Solution to this is follows:
Denote by $A_i$ the set if distribution in which the $ith$ gentlemen gets his umbrellas.
each $|A_i| = 24$
each $|A_i A_j| = 6$
each $|A_i A_j A_k| = 2$
each $|A_i A_j A_k A_l| = 1$
each $|A_1 A_2 A_3 A_4 A_5| = 1$
(sign between these is of intersection)
and hence the number of ways in which none gets his umbrellas is equal to
$|A'_1 A'_2 A'_3 A'_4 A'_5| = 120 - \dbinom{5}{1} \times24 + \dbinom{5}{2} \times6 - \dbinom{5}{3} \times2 + \dbinom{5}{4} \times1 - 1$
What I don't understand is how $|A_i A_j| = 6$ and below this.
And why they had used $\dbinom{5}{1}$ and other such symbols.
$|A_iA_j|=6$ because the $i$th and $j$th gentlemen must receive their own umbrellas, but the other three umbrellas can be permuted in $3!=3\cdot 2 \cdot 1 = 6$ ways among the remaining three gentlemen.
The binomial coefficients are used because the principle of inclusion/exclusion says that you must include all pairwise intersections, three way intersections, etc. It turns out all $n$-way intersections have the same size, which is why you can compute them generally as $|A_iA_j|$ or $|A_iA_jA_k|$, etc. But you also need to count how many such $n$-way intersections there are, which is $\binom{5}{n}$.